Question

Derivation of third equation of motion through graphical method​

Answer 1

Answer:

Explanation:

From the graph, we can say that

The total distance travelled, s is given by the Area of trapezium OABC.

Hence,

S = ½ (Sum of Parallel Sides) × Height

S=(OA+CB)×OC

Since, OA = u, CB = v, and OC = t

The above equation becomes

S= ½ (u+v) × t

Now, since t = (v – u)/ a

The above equation can be written as:

S= ½ ((u+v) × (v-u))/a

Rearranging the equation, we get

S= ½ (v+u) × (v-u)/a

S = (v2-u2)/2a

Third equation of motion is obtained by solving the above equation:

v2 = u2+2aS

Answer 2

Explanation:

Third Equation of Motion by Graphical Method

Consider an object is moving with a uniform acceleration “a” along a straight line. The initial and final velocities of the object at time t = 0 and t = t are u and v respectively. During time t, let s be the total distance travelled by the object. In uniformly acceleration motion the velocity – time graph of an object is a straight line, inclined to the time axis.

OD = u, OC = v and OE = DA = t

Let, the Initial velocity of the object = u

Let, the object is moving with uniform acceleration, a

Let, the object reaches at point B after time t, and its final velocity becomes v.

Draw a line parallel to x-axis DA from point, D from where object starts moving.

Draw another line BA from point B parallel to Y-axis which meets at E at y-axis.

Third Equation of Motion: The Distance covered by the object moving with uniform acceleration is given by the area of trapezium ABDOE.

∴ Area of trapezium ABDOE = ½ x (Sum of Parallel Slides + Distance between Parallel Slides)

Distance (s) = ½ (DO + BE) x OE = ½ (u + v) x t … (3)

Now from equation (2): a=v−ut,

∴ t=v−ua … 4

Now, substitute equation (4) in equation (3) we get:

s=12(u+v)×(v−ua),

s = ½a (v + u) (v – u)

2as = (v + u) (v – u)

2as = v² – u²

v² = u² + 2as

This Expression gives the relation between position and velocity.